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| 1. |
State theorem of perpendicular and parallel axis |
| Answer» Let m and r be the respective masses of the hollow cylinder and the solid sphere.\xa0moment of inertia of the hollow cylinder\xa0I1\xa0= mr2The moment of inertia of the solid sphere I2\xa0{tex}= \\frac { 2 } { 5 } m r ^ { 2 }{/tex}We have the relation:{tex}\\tau = I a{/tex}For the hollow cylinder, {tex}\\tau _ { 1 } = I _ { 1 } \\alpha _ { 1 }{/tex}For the solid sphere, {tex}\\tau _ { 2 } = I _ { 2 } \\alpha _ { 2 }{/tex}As an equal torque is applied to both the bodies, {tex}\\tau _ { 1 } = \\tau _ { 2 }{/tex}{tex}\\therefore \\frac {\\alpha_ { 2 } } { \\alpha _ { 1 } } = \\frac { I _ { 1} } { I _ { 2 } } = \\frac { M r ^ { 2 } } { \\frac { 2 } { 5 } M r ^ { 2 } } = \\frac { 2 } { 5 }{/tex}{tex}a _ { 2 } > a _ { 1 }{/tex} ….(i)Now, using the relation:{tex}\\omega = \\omega _ { 0 } + a t{/tex}{tex}\\omega \\propto a{/tex} …(ii)From equations (i) and (ii), we can write:{tex}\\omega _ { 2 } > \\omega _ { 1 }{/tex}Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder. | |